Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've got the following system of equations:

$$ \begin{cases} x_1'=-8x_1^3-x_2 \\x_2'=-4x_2-4x_1^3 \end{cases} $$

I'm trying to check, if the equilibrium point in $(0,0)$ is stable or not. I am supposed to find so called "Lyapunov function" L, i.e. function which satisfies three following conditions:

1) $L(x_1,x_2)=0$ iff $(x_1,x_2)=(0,0)$

2) $\forall_{(x_1,x_2)\neq(0,0)}L(x_1,x_2)>0$

3) $\forall_{(x_1,x_2)}:\frac{dL}{dt}(x_1,x_2)\leqslant 0$ or $\frac{dL}{dt}(x_1,x_2)\geqslant 0$

The sign in 3) gives us the information about stability in the equilibrium point.

I was trying to find a polynomial $\in\mathbb{R}[x,y]$ with only even exponents, but I was unable to find such function. Is there any way to do it without just "guessing"?

What is the Lyapunov function for this system of ODEs?

share|improve this question
Just to mention $\frac{dL}{dt}=\nabla L\cdot f$, where $f$ is the right hand side of your system. –  ellya Jun 5 '14 at 8:46
Got something from my answer? –  Did Jun 13 '14 at 8:11

2 Answers 2

Hint: If $L(x_1,x_2)$ has a part $x_1^{2n}$, the stability condition involves some multiples of $x_1^{2n+2}$ and $x_1^{2n-1}x_2$. If $L(x_1,x_2)$ has a part $x_2^{2m}$, the stability condition involves some multiples of $x_2^{2m}$ and $x_2^{2m-1}x_1^3$. The mixed terms $x_1^{2n-1}x_2$ and $x_2^{2m-1}x_1^3$ are the same if $(n,m)=(2,1)$.

This suggest to check whether $L(x_1,x_2)=x_1^4+x_2^2$ is a Lyapunov function. If it was not (it is), one could have tried $L(x_1,x_2)=ax_1^4+x_2^2$ for some positive $a$, or even $L(x_1,x_2)=ax_1^4+bx_1^2x_2+x_2^2$ for some $(a,b)$ such that $b^2\lt4a$.

share|improve this answer
This is a nice constructive method for finding a lyapunov function, +1 –  ellya Jun 5 '14 at 8:48

But differentiating your $L$, I get: $$\frac{dL}{dt}(x_1,x_2)=(x_1^4+x_2^2)'=4x_1^3 \cdot x_1'+2x_2\cdot x_2'=4x_1^3\cdot(-8x_1^3-x_2)+2x_2\cdot(-4x_2-4x_1^3)=-32x_1^6-4x_1^3x_2-8x_2^2-8x_1^3x_2=-4\cdot(8x_1^6+3x_1^3x_2+2x_2^2)$$ which doesn't have a constant sign. Differentiating $L=ax_1^4+x_2^2$ works only for $a=-2$, but it is not positive. I am going to try the remaining option.

share|improve this answer
"which doesn't have a constant sign" Really? You might want to reconsider. (That is, if this answer is actually a comment on my answer.) –  Did Jun 5 '14 at 12:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.